### Solution:

t is given that 15 places are occupied. This includes the owner's car also, and, hence 14 other cars are parked.

There are 24 places (excluding places at the two ends) out of which 14 places can be chosen in ${}^{22}{\mathrm{C}}_{14}$ ways.

Excluding 15 (c) 92 (d) none of these the neighboring places there are 22 places in which 14 cars can be

parked in${}^{22}{\mathrm{C}}_{14}$ ways

Hence, required probability $=\frac{{}^{22}{C}_{14}}{{}^{24}{C}_{14}}=\frac{15}{92}$